Problem: Simplify; express your answer in exponential form. Assume $a\neq 0, t\neq 0$. $\dfrac{{(a^{3})^{-3}}}{{(a^{-3}t^{-2})^{-5}}}$
Answer: To start, try working on the numerator and the denominator independently. In the numerator, we have ${a^{3}}$ to the exponent ${-3}$ . Now ${3 \times -3 = -9}$ , so ${(a^{3})^{-3} = a^{-9}}$ In the denominator, we can use the distributive property of exponents. ${(a^{-3}t^{-2})^{-5} = (a^{-3})^{-5}(t^{-2})^{-5}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(a^{3})^{-3}}}{{(a^{-3}t^{-2})^{-5}}} = \dfrac{{a^{-9}}}{{a^{15}t^{10}}}$ Break up the equation by variable and simplify. $\dfrac{{a^{-9}}}{{a^{15}t^{10}}} = \dfrac{{a^{-9}}}{{a^{15}}} \cdot \dfrac{{1}}{{t^{10}}} = a^{{-9} - {15}} \cdot t^{- {10}} = a^{-24}t^{-10}$.